Probability for playing cards

Mr.A has 13 cards of the same suit. He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card, i.e he will take 3 from king whose value is 13, 0 from 10, 9 from 9 and so on.. What is the probability that he can form a number that is divisible by 2?
My working(which is incorrect):

Since this question is dealing with combination. I could answer it as:

Mr.A has 13 cards of the same suit.
He withdraws 4 cards from it and makes a number using the digits in the units place of each chosen card,
That is: .Ace = 1, Deuce = 2, Trey = 3, . . . Ten = 0, Jack = 1, Queen = 2, King = 3.
What is the probability that he can form a number that is divisible by 2?

"Divisible by 2" means that the 4-digit number is even.

There are: 7 odd digits and 6 even digits.

He can make an even number if at least one digit is even.

He will fail of all four digits are odd.. . There are: . ways to get 4 odd digits.. . . . There are: . possible outcomes.. . Hence: .

Soroban's way is the way I would do it but it also easy to see it this way. Being divisible by 2 means it has to be even so out of the 13 cards, there are 6 even cards {2,4,6,8,10,12} . The probability of the last number being even is 6 out of 13 cards. 6/13 !

In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.

The book answer gives

which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.

However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.

In forming a number from the 4 digits,
the even digit may be placed at the end to make it even,
thus forming an even number from the digits available.
You don't need to stick to the order the digits came in.

The book answer gives

which is the probability that the first digit is even,
considering that it doesn't matter whether the others are even or odd.

However, this misses the probabilities of the 1st being odd, the 1st 2 being odd,
the 1st 3 being odd, the 1st and 3rd being odd......etc.

You are calculating the probability if we must take the digitsin the order they come in.

The way the question is worded suggests that Mr. A chooses the 4 cards and tries to make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.

Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.

You are calculating the probability if we must take the digitsin the order they come in.

The way the question is worded suggests that Mr. A chooses the 4 cards and tries to make an even number by placing an even number in the
units position whenever he has the opportunity, in other words by rearranging
the digits deliberately.

Maybe the book did not want him to be allowed do this
and the answer it has given suggests he isn't allowed to.